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Proof step 54 from proof inv-ls-succ (previous step)

nσ[m] = m2, max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ[n0] = n + 1) = ∞, n1 = m2 + 1, ⌜m⌝ < # nσ, 1 + n < n1, stutf(nσ), ∞.n ≤ ((λ m. max∞(λ n. ⌜n⌝ < # nσ ∧ nσ[n] = m).n) to n1)[n], ∀ m. (∃ m0. ⌜m0⌝ < # nσ ∧ nσ[m0] = m - 1 ∧ (# nσ ≠ ∞ ∨ ¬(∀ m1. m0 < m1 → nσ[m1] = m - 1))) → ¬ n1 < m, # nσ ≠ ∞ ∨ ¬(m < m0 → nσ[m0] = m2) ⊦ n1 = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 55:
max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ[n0] = n + 1) = ∞, n < nσ[m], ⌜m⌝ < # nσ, stutf(nσ), ∞.n ≤ max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ[n0] = n).n, ∀ m0. (∃ m1. ⌜m1⌝ < # nσ ∧ nσ[m1] = m0 - 1 ∧ (# nσ ≠ ∞ ∨ ¬(∀ m3. m1 < m3 → nσ[m3] = m0 - 1))) → ¬ nσ[m] + 1 < m0, # nσ ≠ ∞ ∨ ¬(m < m0 → nσ[m0] = nσ[m]) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%m < n ⊦ (nnf to n)[m] = nnf(m); (s)% ⊦ m < n → ¬ n < m; (s)% ⊦ m + n < m + n0 ↔ n < n0; (s)% ⊦ m < n + 1 ↔ ¬ n < m; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m + 1 ≠ 0; (s)